Interference and Diffraction: Coating on a Lens to Minimize Reflection

AI Thread Summary
To minimize reflection of 661 nm light on a lens, a coating with an index of refraction of 1.40 is required, while the glass has an index of 1.65. The minimum thickness of the coating should be calculated to achieve destructive interference, taking into account the 180-degree phase shift that occurs when light reflects off a medium of higher refractive index. The effective wavelength of light in the coating must be considered, as it changes due to the refractive index. The thickness of the coating should be half of the wavelength that results in destructive interference. Understanding the relationship between the thickness, refractive index, and phase changes is crucial for solving this problem.
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Homework Statement


What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 661 nm light? The index of refraction of the coating material is 1.40 and the index of the glass is 1.65.

Hint: You must take into account the 180 degree phase-jump on reflection of light from a medium with a lower index of refraction to a medium of higher index of refraction. Also remember that the wavelength of light changes in a medium. The antireflective coating should be half of the minimum length you determine is necessary for destructive interference to occur.

Homework Equations


Dsin(theta)=m*lambda


The Attempt at a Solution


I don't know how to approach this problem. I have read http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/antiref.html#c3 but I don't see it. Can anyone help me?
 
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You are trying to find the thickness where light reflected off the front surface and light traveling through the coating and reflecting from the rear surface are out of phase - they then interfere destructively and cancel, giving no reflection.
 
That's what I think I have been asked to find, however I cannot come up with an equation. Do you have any other hints? I would like to understand the problem better. Thank you for the quick reply!
 
It's really just a matter of drawing the diagram - you need a whole number of half wavelengths in the thickness of the layer.
Remember that you have two trips through the layer, the 'distance' inside the layer is effected by the refractive index, and you have to be careful about phase changes at the edge
 
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